RBSE Solutions Class 11 Maths Chapter 10 Limits and Derivatives Exercise 10.3

Question 1. Find the derivative of \( x^2 - 2 \) at \( x = 10 \).
Answer: Let \( f(x) = x^2 - 2 \).
We need to find the derivative at \( x = 10 \).
\[ f'(10) = \lim_{h \to 0} \frac{f(10+h) - f(10)}{h} \]
\[ = \lim_{h \to 0} \frac{((10+h)^2 - 2) - (10^2 - 2)}{h} \]
\[ = \lim_{h \to 0} \frac{(10+h)^2 - 2 - 10^2 + 2}{h} \]
\[ = \lim_{h \to 0} \frac{(10+h)^2 - 10^2}{h} \]
Using the identity \( a^2 - b^2 = (a+b)(a-b) \):
\[ = \lim_{h \to 0} \frac{(10+h+10)(10+h-10)}{h} \]
\[ = \lim_{h \to 0} \frac{(20+h)(h)}{h} \]
\[ = \lim_{h \to 0} (20+h) \]
Now, substitute \( h = 0 \):
\[ = 20 + 0 = 20 \]
So, the derivative of \( x^2 - 2 \) at \( x = 10 \) is \( 20 \).
In simple words: To find the derivative, we use the first principle formula. We substitute the function and \( x=10 \) into the formula, simplify it, and then plug in \( h=0 \) to get the final answer.

🎯 Exam Tip: Remember the first principle of differentiation and the algebraic identity \( a^2 - b^2 = (a-b)(a+b) \) to simplify expressions efficiently.

Question 3. Find the derivative of the following function from first principle:
(i) \( x^3 - 16 \)
(ii) \( (x - 1) (x - 2) \)
(iii) \( \frac { 1 }{ x^2 } \)
(iv) \( \frac { x+1 }{ x-1 } \)
Answer:
(i) Let \( y = x^3 - 16 \).
Let \( y + \delta y = (x + \delta x)^3 - 16 \).
This means \( \delta y = (x + \delta x)^3 - 16 - y \).
Substitute \( y = x^3 - 16 \):
\( \delta y = (x + \delta x)^3 - 16 - (x^3 - 16) \)
\( \delta y = (x + \delta x)^3 - x^3 \)
Using the identity \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \), where \( a = (x + \delta x) \) and \( b = x \):
\( \delta y = (x + \delta x - x)((x + \delta x)^2 + x(x + \delta x) + x^2) \)
\( \delta y = \delta x ((x + \delta x)^2 + x(x + \delta x) + x^2) \)
Now, divide by \( \delta x \):
\( \frac{\delta y}{\delta x} = (x + \delta x)^2 + x(x + \delta x) + x^2 \)
Next, take the limit as \( \delta x \to 0 \):
\[ \frac{dy}{dx} = \lim_{\delta x \to 0} \left( (x + \delta x)^2 + x(x + \delta x) + x^2 \right) \]
\[ = (x + 0)^2 + x(x + 0) + x^2 \]
\[ = x^2 + x^2 + x^2 \]
\[ = 3x^2 \]
So, \( \frac{d}{dx}(x^3 - 16) = 3x^2 \).

(ii) Let \( y = (x - 1) (x - 2) \).
First, expand the expression: \( y = x^2 - 2x - x + 2 = x^2 - 3x + 2 \).
Let \( y + \delta y = (x + \delta x)^2 - 3(x + \delta x) + 2 \).
This means \( \delta y = (x + \delta x)^2 - 3(x + \delta x) + 2 - y \).
Substitute \( y = x^2 - 3x + 2 \):
\( \delta y = (x + \delta x)^2 - 3(x + \delta x) + 2 - (x^2 - 3x + 2) \)
\( \delta y = (x + \delta x)^2 - x^2 - 3(x + \delta x) + 3x \)
Factor out common terms:
\( \delta y = [(x + \delta x)^2 - x^2] - [3(x + \delta x) - 3x] \)
Using \( a^2 - b^2 = (a-b)(a+b) \) for the first term and factoring 3 from the second:
\( \delta y = (x + \delta x - x)(x + \delta x + x) - 3(x + \delta x - x) \)
\( \delta y = \delta x (2x + \delta x) - 3 \delta x \)
Now, divide by \( \delta x \):
\( \frac{\delta y}{\delta x} = (2x + \delta x) - 3 \)
Next, take the limit as \( \delta x \to 0 \):
\[ \frac{dy}{dx} = \lim_{\delta x \to 0} (2x - 3 + \delta x) \]
\[ = 2x - 3 + 0 \]
\[ = 2x - 3 \]
So, \( \frac{d}{dx}((x - 1)(x - 2)) = 2x - 3 \).

(iii) Let \( y = \frac{1}{x^2} = x^{-2} \).
Let \( y + \delta y = (x + \delta x)^{-2} \).
\( \delta y = (x + \delta x)^{-2} - x^{-2} \)
\( \delta y = \frac{1}{(x + \delta x)^2} - \frac{1}{x^2} \)
Find a common denominator:
\( \delta y = \frac{x^2 - (x + \delta x)^2}{x^2 (x + \delta x)^2} \)
Using \( a^2 - b^2 = (a-b)(a+b) \) in the numerator:
\( \delta y = \frac{(x - (x + \delta x))(x + (x + \delta x))}{x^2 (x + \delta x)^2} \)
\( \delta y = \frac{(-\delta x)(2x + \delta x)}{x^2 (x + \delta x)^2} \)
Now, divide by \( \delta x \):
\( \frac{\delta y}{\delta x} = \frac{-(2x + \delta x)}{x^2 (x + \delta x)^2} \)
Next, take the limit as \( \delta x \to 0 \):
\[ \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{-(2x + \delta x)}{x^2 (x + \delta x)^2} \]
Substitute \( \delta x = 0 \):
\[ = \frac{-(2x + 0)}{x^2 (x + 0)^2} \]
\[ = \frac{-2x}{x^2 \cdot x^2} \]
\[ = \frac{-2x}{x^4} \]
\[ = -\frac{2}{x^3} \]
So, \( \frac{d}{dx}(\frac{1}{x^2}) = -\frac{2}{x^3} \).

(iv) Let \( y = \frac{x+1}{x-1} \).
Let \( y + \delta y = \frac{(x + \delta x)+1}{(x + \delta x)-1} \).
\( \delta y = \frac{x + \delta x + 1}{x + \delta x - 1} - \frac{x+1}{x-1} \)
Find a common denominator:
\( \delta y = \frac{(x + \delta x + 1)(x-1) - (x+1)(x + \delta x - 1)}{((x + \delta x - 1)(x-1))} \)
Expand the numerator:
\( (x^2 - x + x\delta x - \delta x + x - 1) - (x^2 + x\delta x - x + x + \delta x - 1) \)
\( (x^2 + x\delta x - \delta x - 1) - (x^2 + x\delta x + \delta x - 1) \)
\( x^2 + x\delta x - \delta x - 1 - x^2 - x\delta x - \delta x + 1 \)
\( = -2\delta x \)
So, \( \delta y = \frac{-2\delta x}{((x + \delta x - 1)(x-1))} \)
Now, divide by \( \delta x \):
\( \frac{\delta y}{\delta x} = \frac{-2}{(x + \delta x - 1)(x-1)} \)
Next, take the limit as \( \delta x \to 0 \):
\[ \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{-2}{(x + \delta x - 1)(x-1)} \]
Substitute \( \delta x = 0 \):
\[ = \frac{-2}{(x + 0 - 1)(x-1)} \]
\[ = \frac{-2}{(x-1)(x-1)} \]
\[ = \frac{-2}{(x-1)^2} \]
So, \( \frac{d}{dx}(\frac{x+1}{x-1}) = \frac{-2}{(x-1)^2} \).
In simple words: For each function, we use the definition of the derivative (first principle). This means finding the change in \( y \) for a small change in \( x \), dividing by the change in \( x \), and then taking the limit as the change in \( x \) becomes zero. Simplify the expressions using algebraic rules at each step.

🎯 Exam Tip: When using the first principle, carefully expand and simplify algebraic terms. Pay close attention to signs and use common identities like \( a^2 - b^2 \) or \( a^3 - b^3 \) to simplify the numerator before taking the limit.

Question 4. For the function \( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + ... + \frac{x^2}{2} + x + 1 \) Prove that \( f'(1) = 100 f'(0) \).
Answer: Given the function \( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + ... + \frac{x^2}{2} + x + 1 \).
First, find the derivative \( f'(x) \). We use the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \):
\[ f'(x) = \frac{d}{dx}\left(\frac{x^{100}}{100}\right) + \frac{d}{dx}\left(\frac{x^{99}}{99}\right) + ... + \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}(x) + \frac{d}{dx}(1) \]
\[ f'(x) = \frac{1}{100} (100x^{99}) + \frac{1}{99} (99x^{98}) + ... + \frac{1}{2} (2x) + 1 + 0 \]
\[ f'(x) = x^{99} + x^{98} + ... + x + 1 \]
Now, substitute \( x = 1 \) into \( f'(x) \):
\[ f'(1) = 1^{99} + 1^{98} + ... + 1 + 1 \]
This is a sum of \( 100 \) ones (from \( x^0 \) to \( x^{99} \)).
So, \( f'(1) = 100 \).
Next, substitute \( x = 0 \) into \( f'(x) \):
\[ f'(0) = 0^{99} + 0^{98} + ... + 0 + 1 \]
\[ f'(0) = 1 \]
We need to prove that \( f'(1) = 100 f'(0) \).
Substitute the values we found:
\( 100 = 100 \times 1 \)
\( 100 = 100 \)
This proves the statement.
In simple words: First, find the derivative of the given function. Then, put \( x=1 \) into the derivative to get the value of \( f'(1) \). After that, put \( x=0 \) into the derivative to find \( f'(0) \). Finally, check if \( f'(1) \) is 100 times \( f'(0) \).

🎯 Exam Tip: Remember to differentiate each term correctly using the power rule. When evaluating at \( x=0 \), be careful with terms like \( x^0 \) which equals 1.

Question 6. For some constant a and b, find the derivative of the following functions :
(i) \( (x - a) (x - b) \)
(ii) \( (ax^2 + b)^2 \)
(iii) \( \frac{x-a}{x-b} \)
Answer:
(i) Let \( y = f(x) = (x - a) (x - b) \).
Expand the expression first: \( y = x^2 - bx - ax + ab = x^2 - (a + b)x + ab \).
Now, differentiate with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab) \]
\[ = 2x - (a+b) \cdot 1 + 0 \]
\[ = 2x - a - b \]
So, the derivative of \( (x - a) (x - b) \) is \( 2x - a - b \).

(ii) Let \( y = f(x) = (ax^2 + b)^2 \).
Expand the expression first: \( y = (ax^2)^2 + 2(ax^2)(b) + b^2 = a^2x^4 + 2abx^2 + b^2 \).
Now, differentiate with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + \frac{d}{dx}(b^2) \]
\[ = a^2 \cdot (4x^{4-1}) + 2ab \cdot (2x^{2-1}) + 0 \]
\[ = 4a^2x^3 + 4abx \]
We can factor out \( 4ax \) from the expression:
\[ = 4ax(ax^2 + b) \]
So, the derivative of \( (ax^2 + b)^2 \) is \( 4a^2x^3 + 4abx \) or \( 4ax(ax^2 + b) \).

(iii) Let \( y = f(x) = \frac{x-a}{x-b} \).
We use the quotient rule for differentiation: If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, \( u = x - a \) and \( v = x - b \).
So, \( \frac{du}{dx} = \frac{d}{dx}(x-a) = 1 - 0 = 1 \).
And \( \frac{dv}{dx} = \frac{d}{dx}(x-b) = 1 - 0 = 1 \).
Substitute these into the quotient rule formula:
\[ \frac{dy}{dx} = \frac{(x-b) \cdot (1) - (x-a) \cdot (1)}{(x-b)^2} \]
\[ = \frac{x-b - (x-a)}{(x-b)^2} \]
\[ = \frac{x-b - x + a}{(x-b)^2} \]
\[ = \frac{a-b}{(x-b)^2} \]
So, the derivative of \( \frac{x-a}{x-b} \) is \( \frac{a-b}{(x-b)^2} \).
In simple words: For part (i) and (ii), expand the functions first and then apply the power rule to differentiate each term. For part (iii), use the quotient rule, which helps differentiate functions written as one expression divided by another. Make sure to simplify the algebra after applying the rules.

🎯 Exam Tip: For products and quotients, it's often easier to expand first if possible before differentiating term-by-term. If expansion is too complex, use the product rule or quotient rule. Clearly state which rule you are using.

Question 7. For any constant a, find the derivative of \( \frac{x^n - a^n}{x - a} \)
Answer: Let \( y = \frac{x^n - a^n}{x - a} \).
We use the quotient rule: If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, \( u = x^n - a^n \) and \( v = x - a \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{du}{dx} = \frac{d}{dx}(x^n - a^n) = nx^{n-1} - 0 = nx^{n-1} \) (since \( a^n \) is a constant).
\( \frac{dv}{dx} = \frac{d}{dx}(x - a) = 1 - 0 = 1 \).
Now, apply the quotient rule:
\[ \frac{dy}{dx} = \frac{(x-a)(nx^{n-1}) - (x^n - a^n)(1)}{(x-a)^2} \]
Expand the numerator:
\[ = \frac{nx^{n-1} \cdot x - nx^{n-1} \cdot a - (x^n - a^n)}{(x-a)^2} \]
\[ = \frac{nx^n - nax^{n-1} - x^n + a^n}{(x-a)^2} \]
Combine terms with \( x^n \):
\[ = \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2} \]
So, the derivative of \( \frac{x^n - a^n}{x - a} \) is \( \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2} \).
In simple words: We need to find the derivative of a fraction. We use the quotient rule, which involves finding the derivative of the top and bottom parts separately, then combining them with a specific formula. We carefully substitute and simplify the terms to get the final answer.

🎯 Exam Tip: Be very careful with the algebra when applying the quotient rule, especially when dealing with exponents like \( x^n \) and constants like \( a^n \). Remember that the derivative of a constant is zero.

Question 8. Find the derivative of the following 3
(i) \( 2x - \frac{3}{4} \)
(ii) \( (5x^3 + 3x - 1) (x - 1) \)
(iii) \( x^5 (3 - 6x - 9) \)
(iv) \( x^{-4} (3 - 4x - 5) \)
(v) \( \frac{2}{x+1} + \frac{x^2}{3x-1} \)
Answer:
(i) Let \( f(x) = 2x - \frac{3}{4} \).
Differentiate with respect to \( x \):
\[ \frac{d}{dx} \left(2x - \frac{3}{4}\right) = \frac{d}{dx}(2x) - \frac{d}{dx}\left(\frac{3}{4}\right) \]
\[ = 2 \cdot 1 - 0 \]
\[ = 2 \]
So, the derivative of \( 2x - \frac{3}{4} \) is \( 2 \).

(ii) Let \( f(x) = (5x^3 + 3x - 1) (x - 1) \).
This is a product of two functions. Let \( u = 5x^3 + 3x - 1 \) and \( v = x - 1 \).
We use the product rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{du}{dx} = \frac{d}{dx}(5x^3 + 3x - 1) = 5(3x^2) + 3(1) - 0 = 15x^2 + 3 \).
\( \frac{dv}{dx} = \frac{d}{dx}(x - 1) = 1 - 0 = 1 \).
Now, apply the product rule:
\[ \frac{d}{dx}(f(x)) = (5x^3 + 3x - 1)(1) + (x - 1)(15x^2 + 3) \]
\[ = 5x^3 + 3x - 1 + (x \cdot 15x^2 + x \cdot 3 - 1 \cdot 15x^2 - 1 \cdot 3) \]
\[ = 5x^3 + 3x - 1 + 15x^3 + 3x - 15x^2 - 3 \]
Combine like terms:
\[ = (5x^3 + 15x^3) - 15x^2 + (3x + 3x) - 1 - 3 \]
\[ = 20x^3 - 15x^2 + 6x - 4 \]
So, the derivative of \( (5x^3 + 3x - 1) (x - 1) \) is \( 20x^3 - 15x^2 + 6x - 4 \).

(iii) Let \( f(x) = x^5 (3 - 6x - 9) \).
First, simplify the expression: \( f(x) = x^5 (-6x - 6) = -6x^6 - 6x^5 \).
Now, differentiate with respect to \( x \):
\[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(-6x^6) - \frac{d}{dx}(6x^5) \]
\[ = -6(6x^{6-1}) - 6(5x^{5-1}) \]
\[ = -36x^5 - 30x^4 \]
So, the derivative of \( x^5 (3 - 6x - 9) \) is \( -36x^5 - 30x^4 \).

(iv) Let \( f(x) = x^{-4} (3 - 4x - 5) \).
First, simplify the expression: \( f(x) = x^{-4} (-4x - 2) = -4x^{-3} - 2x^{-4} \).
Now, differentiate with respect to \( x \):
\[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(-4x^{-3}) - \frac{d}{dx}(2x^{-4}) \]
\[ = -4(-3x^{-3-1}) - 2(-4x^{-4-1}) \]
\[ = 12x^{-4} + 8x^{-5} \]
This can also be written as \( \frac{12}{x^4} + \frac{8}{x^5} \).
So, the derivative of \( x^{-4} (3 - 4x - 5) \) is \( \frac{12}{x^4} + \frac{8}{x^5} \).

(v) Let \( f(x) = \frac{2}{x+1} + \frac{x^2}{3x-1} \).
We differentiate each term separately.
For the first term, \( \frac{d}{dx}\left(\frac{2}{x+1}\right) \). Let \( u=2, v=x+1 \). Then \( \frac{du}{dx}=0, \frac{dv}{dx}=1 \).
Using the quotient rule: \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{(x+1)(0) - 2(1)}{(x+1)^2} = \frac{-2}{(x+1)^2} \).
For the second term, \( \frac{d}{dx}\left(\frac{x^2}{3x-1}\right) \). Let \( u=x^2, v=3x-1 \). Then \( \frac{du}{dx}=2x, \frac{dv}{dx}=3 \).
Using the quotient rule: \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{(3x-1)(2x) - x^2(3)}{(3x-1)^2} \)
\[ = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2} \]
Combine the derivatives of both terms:
\[ f'(x) = \frac{-2}{(x+1)^2} + \frac{3x^2 - 2x}{(3x-1)^2} \]
So, the derivative of \( \frac{2}{x+1} + \frac{x^2}{3x-1} \) is \( \frac{-2}{(x+1)^2} + \frac{3x^2 - 2x}{(3x-1)^2} \).
In simple words: For functions that are sums or differences, differentiate each part separately. If a function is a product, use the product rule. If it's a fraction, use the quotient rule. Remember to simplify the expression first if possible, like in parts (iii) and (iv), before differentiating.

🎯 Exam Tip: Always simplify polynomial expressions before differentiating to avoid unnecessary use of product rule. For fractions, the quotient rule is essential, but remember that constants in the numerator can simplify the process.

Question 9. Find the derivative of cos x by first principle.
Answer: Let \( f(x) = \cos x \).
By the first principle, the derivative \( f'(x) \) is:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
\[ = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \]
Use the trigonometric identity \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \).
Here, \( A = x+h \) and \( B = x \).
\( \frac{A+B}{2} = \frac{x+h+x}{2} = \frac{2x+h}{2} \)
\( \frac{A-B}{2} = \frac{x+h-x}{2} = \frac{h}{2} \)
Substitute these into the limit expression:
\[ = \lim_{h \to 0} \frac{-2 \sin\left(\frac{2x+h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} \]
Rearrange the terms to use the limit \( \lim_{k \to 0} \frac{\sin k}{k} = 1 \):
\[ = \lim_{h \to 0} \left( -\sin\left(\frac{2x+h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \]
Now, take the limit. As \( h \to 0 \), \( \frac{h}{2} \to 0 \):
\[ = -\sin\left(\frac{2x+0}{2}\right) \cdot \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \]
\[ = -\sin(x) \cdot 1 \]
\[ = -\sin x \]
So, the derivative of \( \cos x \) is \( -\sin x \).
In simple words: To find the derivative of cosine x, we use the first principle. We substitute the cosine function into the formula and use a special trigonometry rule to change the subtraction of cosines into a product of sines. Then, we arrange the terms to use a known limit for \( \sin k / k \), which helps us find that the derivative is minus sine x.

🎯 Exam Tip: Mastering trigonometric identities, especially for sums and differences of sines and cosines, is crucial for first-principle differentiation of trigonometric functions. Remember the key limit: \( \lim_{k \to 0} \frac{\sin k}{k} = 1 \).

Question 10. Find the derivatives of the following :
(i) sin x cos x
(ii) sec x
(iii) cosec x
(iv) 3 cot x + 5 cosec x
(v) 5 sin x - 6 cos x + 7
Answer:
(i) Let \( f(x) = \sin x \cos x \).
We use the product rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
Here, \( u = \sin x \) and \( v = \cos x \).
\( \frac{du}{dx} = \cos x \) and \( \frac{dv}{dx} = -\sin x \).
So, \( f'(x) = \sin x (-\sin x) + \cos x (\cos x) \)
\[ = -\sin^2 x + \cos^2 x \]
Using the trigonometric identity \( \cos^2 x - \sin^2 x = \cos 2x \):
\[ = \cos 2x \]
So, the derivative of \( \sin x \cos x \) is \( \cos 2x \).

(ii) Let \( f(x) = \sec x \).
We can write \( \sec x = \frac{1}{\cos x} \). We use the quotient rule. Let \( u=1, v=\cos x \).
\( \frac{du}{dx} = 0 \) and \( \frac{dv}{dx} = -\sin x \).
\[ f'(x) = \frac{\cos x (0) - 1 (-\sin x)}{(\cos x)^2} \]
\[ = \frac{\sin x}{\cos^2 x} \]
\[ = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \]
\[ = \tan x \sec x \]
So, the derivative of \( \sec x \) is \( \sec x \tan x \).

(iii) Let \( f(x) = \operatorname{cosec} x \).
We can write \( \operatorname{cosec} x = \frac{1}{\sin x} \). We use the quotient rule. Let \( u=1, v=\sin x \).
\( \frac{du}{dx} = 0 \) and \( \frac{dv}{dx} = \cos x \).
\[ f'(x) = \frac{\sin x (0) - 1 (\cos x)}{(\sin x)^2} \]
\[ = \frac{-\cos x}{\sin^2 x} \]
\[ = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \]
\[ = -\cot x \operatorname{cosec} x \]
So, the derivative of \( \operatorname{cosec} x \) is \( -\operatorname{cosec} x \cot x \).

(iv) Let \( f(x) = 3 \cot x + 5 \operatorname{cosec} x \).
Differentiate each term separately:
\[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(3 \cot x) + \frac{d}{dx}(5 \operatorname{cosec} x) \]
We know that \( \frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x \) and \( \frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x \).
\[ = 3(-\operatorname{cosec}^2 x) + 5(-\operatorname{cosec} x \cot x) \]
\[ = -3 \operatorname{cosec}^2 x - 5 \operatorname{cosec} x \cot x \]
So, the derivative of \( 3 \cot x + 5 \operatorname{cosec} x \) is \( -3 \operatorname{cosec}^2 x - 5 \operatorname{cosec} x \cot x \).

(v) Let \( f(x) = 5 \sin x - 6 \cos x + 7 \).
Differentiate each term separately:
\[ \frac{d}{dx}(f(x)) = \frac{d}{dx}(5 \sin x) - \frac{d}{dx}(6 \cos x) + \frac{d}{dx}(7) \]
We know that \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \). The derivative of a constant is 0.
\[ = 5(\cos x) - 6(-\sin x) + 0 \]
\[ = 5 \cos x + 6 \sin x \]
So, the derivative of \( 5 \sin x - 6 \cos x + 7 \) is \( 5 \cos x + 6 \sin x \).
In simple words: For functions involving products of trigonometric terms, use the product rule. For functions that are ratios, use the quotient rule or rewrite them in terms of sine and cosine before applying the quotient rule. For sums and differences, differentiate each term separately using standard derivative formulas for trigonometric functions and constants.

🎯 Exam Tip: Memorize the derivatives of common trigonometric functions (\(\sin x\), \(\cos x\), \(\tan x\), \(\sec x\), \(\operatorname{cosec} x\), \(\cot x\)) and know when to apply the product or quotient rule effectively. Always simplify your final answer using trigonometric identities if possible.